Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in PYTHON):

Solution:
import sys

N=input()
P=map(float,  sys.stdin.readline().split() )

dp=[ [ 0 for j in range(N+1) ] for i in range(N+1) ]
dp[0][0]=1


for i in range(N):
	p=P[i]
	for j in range(i+2):
		if j==0:
			#print "dp[",i+1 , "][0]=dp[", i ,"][0]*(",1-p,")"
			dp[i+1][0]=dp[i][0]*(1-p)
		else:
			#print "dp[",i+1,"][",j,"]=dp[",i,"][",j, "]*(", 1-p,")+dp[",i," ][", j-1," ]*",p
			dp[i+1][j]=dp[i][j]*(1-p)+dp[i][j-1]*p

n=N/2+1

ans=0
for i in range(n, N+1):
	ans+=dp[-1][i]
print ans